Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$

Question

Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,

$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$ Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have

$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\ \int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\ \int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\ &=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\ &=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\ &=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\ &=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare \end{aligned} $$ Furthermore, $$ \begin{aligned} \int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\ &=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\ &=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3) \end{aligned} $$ and $$ \int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3) $$

Answer 1

Method. $(1)$, Use Mittag-Leffler's series for $\cot(x)$

$$\begin{align} I&=\int_0^\frac{\pi}{2}x\ln(\sin(x))~dx=\frac{1}{2}\int_0^\frac{\pi}{2}\ln(\sin(x))~dx^2\\ \\ &=-\frac{1}2 \int_0^\frac{\pi}{2}\frac{x^2}{\tan(x)}~dx=-\frac{1}2\int_0^\frac{\pi}{2}x^2\left(\frac{1}x+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2} \right)~dx \end{align}$$

Next, integrate them term by term and you will get the answer.

Method. $(2)$, Construct two bases, as mentioned by @Quanto:

$$\begin{align} I_1&=\int_0^\frac{\pi}{2}x\left[\ln(\sin(x))+\ln(\cos(x))\right]dx=\int_0^\frac{\pi}{2}x\ln\left( \frac{1}{2}\sin(2x) \right)dx\\ \\ I_2&=\int_0^\frac{\pi}{2}x\left[\ln(\sin(x))-\ln(\cos(x))\right]dx=\int_0^\frac{\pi}{2}x\ln(\tan(x))dx \end{align}$$

Answer 2

Even if there is an antiderivative (have a look here), you could write $$\log(\sin(x))=\log(x)-\sum_{n=1}^\infty(-1)^{n+1}\, \frac{ 2^{2 n-1}\, B_{2 n} }{n \,(2 n)!}x^{2 n}$$ where appear Bernoulli numbers.

Integrate termwise

$$\int_0^{\frac \pi 2}x\,\log(x)=-\frac{1}{16} \pi ^2 (1+2 \log (2)-2 \log (\pi ))$$ $$\int_0^{\frac \pi 2} (-1)^{n+1}\, \frac{ 2^{2 n-1}\, B_{2 n} }{n \,(2 n)!}x^{2 n+1}=(-1)^{n+1}\, \frac{\pi ^{2 n+2} B_{2 n}}{16 n (n+1) (2 n)!}$$ The infinite summation gives $$-\frac{1}{16} \left(7 \zeta (3)+\pi ^2-2 \pi ^2 \log (\pi )\right)$$ and then the result (with a sign difference)

Answer 3

Enforcing the substitution $x\mapsto\frac {\pi}2-x$, then we get

\begin{align*} \int\limits_0^{\pi/2}x\log\sin x\,\mathrm dx & =\int\limits_0^{\pi/2}\left(\frac {\pi}2-x\right)\log\cos x\,\mathrm dx\\ & =\frac {\pi}2\int\limits_0^{\pi/2}\log\cos x\,\mathrm dx-\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx\\ & =-\frac {\pi^2}4\log 2-\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx \end{align*}

Now, express $\cos x$ as

$$2\cos x=e^{ix}+e^{-ix}$$

And expand the logarithm into three separate terms. Since the integral is entirely real, the complex components will cancel out in the end.

\begin{align*} \int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx & =\int\limits_0^{\pi/2}x\log\left(e^{ix}+e^{-ix}\right)\,\mathrm dx-\log 2\int\limits_0^{\pi/2}x\,\mathrm dx\\ & =i\int\limits_0^{\pi/2}x^2\,\mathrm dx+\int\limits_0^{\pi/2}x\log\left(1+e^{-2ix}\right)\,\mathrm dx-\frac {\pi^2}8\log 2\\ & =\frac {\pi^3i}{24}-\frac {\pi^2}8\log 2+\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}n\int\limits_0^{\pi/2}xe^{-2nix}\,\mathrm dx \end{align*}

Focusing on the remaining sum, we use integration by parts

\begin{align*} \sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}n\int\limits_0^{\pi/2}xe^{-2nix}\,\mathrm dx & =-\frac 14\sum\limits_{n=1}^{+\infty}\frac 1{n^3}-\frac {\pi i}4\sum\limits_{n=1}^{+\infty}\frac 1{n^2}-\frac 14\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}{n^3}\\ & =-\frac 7{16}\zeta(3)-\frac {\pi^3 i}{24} \end{align*}

So the integral is

\begin{align*} \int\limits_0^{\pi/2}x\log\sin x\,\mathrm dx=-\frac {\pi^2}4\log 2+\frac {\pi^2}8\log 2+\frac 7{16}\zeta(3)\color{blue}{=-\frac {\pi^2}8\log 2+\frac 7{16}\zeta(3)} \end{align*}

Answer 4

\begin{align}J&=\int_0^{\frac{\pi}{2}}x\ln(\sin x)dx\\ K&=\int_0^{\frac{\pi}{2}}x\ln(\cos x)dx\\ J+K&=\int_0^{\frac{\pi}{2}}x\ln(\sin x\cos x)dx\\ &\overset{y=\frac{\pi}{2}-x}=\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-y\right)\ln(\sin y\cos y)dy\\ &\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln(\sin y\cos y)dy\\ &=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin(2y)}{2}\right)dy\\ &\frac{\pi}{4}\underbrace{\int_0^{\frac{\pi}{2}}\ln\left(\sin(2y)\right)dy}_{z=2y}-\frac{\pi^2}{8}\ln 2\\ &=\frac{\pi}{8}\int_0^\pi \ln(\sin z)dz-\frac{\pi^2}{8}\ln 2\\ &=\frac{\pi}{8}\int_0^{\frac{\pi}{2}} \ln(\sin z)dz+\frac{\pi}{8}\underbrace{\int_{\frac{\pi}{2}}^\pi \ln(\sin z)dz}_{x=\pi-z}-\frac{\pi}{8}\ln 2\\ &=\frac{\pi}{4}\underbrace{\int_0^{\frac{\pi}{2}} \ln(\sin x)dx}_{=-\frac{\pi}{2}\ln 2}-\frac{\pi^2}{8}\ln 2=\boxed{-\frac{\pi^2}{4}\ln 2}\\ J-K&=\int_0^{\frac{\pi}{2}}x\ln(\tan x)dx\\ &\overset{t=\tan x}=\int_0^\infty \frac{\ln t\arctan t}{1+t^2}dt \end{align}

And see:

$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$

https://math.stackexchange.com/a/4500094/186817

Answer 5

Here's a direct attack on the integral

$$i=\int_0^{\frac{\pi }{2}} x \log (\sin (x)) \, dx$$

using only Euler's formula $e^{i x} = \cos(x)+ i \sin(x)$, properties of the $\log$-function $\log(x y)=\log(x) + \log(y)$ including the series expansion $-\log(1-x)=\sum_{k=1}^{\infty}\frac{x^k}{k}$, and the definition of Riemann's zeta function $\zeta(s) = \sum_{k=1}^{\infty}\frac{1}{k^s}$, also the derivative of an integral with respect to a parameter is used. The solution is lengthy because it is elementary and detailed. No tricky formulas or external references are used. Notice that parts of this solution are already contained in other solutions here.

We start by writing

$$\begin{align} \log (\sin (x))& =\log \left(\frac{e^{i x}-e^{-i x}}{2 i}\right)\\ & =\log \left(\frac{e^{i x} \left(1-e^{-2 i x}\right)}{2 i}\right)\\ & =i x+\log (\frac{1}{2 i})+\log \left(1-e^{-2 i x}\right)\end{align}\tag{1}$$

Hence the integral can be written as $i=i_1+i_2$ with

$$\begin{align} i_1 & = \int_0^{\frac{\pi }{2}} x \left(i x+\log (\frac{1}{2 i})\right)\,dx\\ & =-\frac{1}{8} \pi ^2 \log (2)-\frac{1}{48} \left(i \pi ^3\right)\\ i_2 & = \int_0^{\frac{\pi }{2}} x \log \left(1-e^{-2 i x}\right) \, dx\end{align}\tag{2}$$

Now

$$\begin{align} i_2 &=\int_0^{\frac{\pi }{2}} x \log \left(1-e^{-2 i x}\right) \, dx\\ & =-\int_0^{\frac{\pi }{2}} \sum _{k=1}^{\infty } \frac{x e^{-2 i k x}}{k} \, dx\\ & =-\sum _{k=1}^{\infty } \int_0^{\frac{\pi }{2}} \frac{x e^{-2 i k x}}{k} \, dx\\ & = -\sum _{k=1}^{\infty } \frac{1}{k} \left(\frac{1}{-2 i k}\right)\frac{\partial}{\partial{k}}\int_0^{\frac{\pi }{2}} e^{-2 i k x} \, dx\\ & =-\sum _{k=1}^{\infty } \frac{-1+(-1)^k (1+i \pi k)}{4 k^3}\\ & =\frac{1}{4} \sum _{k=1}^{\infty } \frac{1-(-1)^k}{k^3}-\frac{1}{4} (i \pi ) \sum _{k=1}^{\infty } \frac{(-1)^k}{k^2}\\ & =\frac{1}{4} s_1-\frac{i \pi}{4}s_2 \end{align}\tag{3}$$

The first sum is

$$\begin{align} s_1 & =\sum _{k=1}^{\infty } \frac{1-(-1)^k}{k^3}=2 \sum _{k=1}^{\infty } \frac{1}{(2 k-1)^3}\\ & =2 \left(\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+\dots\right)\\ & = 2\left(\left( \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\dots\right)-\left(\frac{1}{2^3}+ \frac{1}{4^3}+\frac{1}{6^3} +\dots\right)\right)\\ & = 2\left(\left( \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\dots\right)-\frac{1}{2^{3}}\left(\frac{1}{1^3}+ \frac{1}{2^3}+\frac{1}{3^3} +\dots\right)\right)\\ & = 2 \left(\zeta(3) -\frac{1}{8}\zeta(3)\right)=\frac{7}{4} \zeta(3) \end{align}\tag{4}$$

and the second sum, similarly,

$$\begin{align} s_2 &=\sum _{k=1}^{\infty } \frac{(-1)^k}{k^2}=\sum _{k=1}^{\infty } \left(\frac{(-1)^k}{k^2}+\frac{1}{k^2}\right)-\sum _{k=1}^{\infty } \frac{1}{k^2}\\ & =2 \sum _{k=1}^{\infty } \frac{1}{(2 k)^2}-\sum _{k=1}^{\infty } \frac{1}{k^2}=\left(\frac{1}{2}-1\right) \zeta (2)\\ & =-\frac{\zeta (2)}{2}=-\frac{\pi ^2}{12}\end{align}\tag{5}$$

So that

$$\begin{align} i_1 & =-\frac{1}{8} \pi ^2 \log (2)-\frac{i \pi^3 }{48}\\ i_2 & =\frac{1}{4}\left(\frac{7}{4} \zeta (3)\right)-\frac{i \pi }{4}\left(-\frac{\pi ^2}{12}\right) =\frac{7}{16} \zeta (3) +\frac{i \pi^3 }{48} \end{align}\tag{6}$$

and finally

$$i=i_1+i_2=-\frac{1}{8} \pi ^2 \log (2)+\frac{7 }{16}\zeta (3)\tag{7}$$

QED.